Is the limit of the spectral radius the spectral radius of the limit?

The answer is no.

This example is due to Kakutani, found in C. E. Rickart's General Theory of Banach Algebras, page $282$.

Consider the Banach space $\ell^2$ with the canonical basis $(e_n)_n$. Define a sequence of scalars $(\alpha_n)_n$ with the relation $\alpha_{2^k(2l+1)} = e^{-k}$ for $k,l \ge 0$.

Define $A : \ell^2 \to \ell^2$ with $Ae_n = \alpha_n e_{n+1}$. We have $\|A\| = \sup_{n\in\mathbb{N}}|\alpha_n|$. Also define a sequence of operators $A_k : \ell^2 \to \ell^2$ with $$A_k e_n = \begin{cases} 0, &\text{ if } n = 2^k(2l+1) \text{ for some } l \ge 0 \\ \alpha_ne_{n+1}, &\text{ if } n \ne 2^k(2l+1) \text{ for some } l \ge 0 \end{cases}$$ Then $A_k^{2^{k+1}} = 0$ so $A_k$ is nilpotent. We also have $A_k \to A$ since $$(A - A_k)e_n = \begin{cases} e^{-k}, &\text{ if } n = 2^k(2l+1) \text{ for some } l \ge 0 \\ 0, &\text{ if } n \ne 2^k(2l+1) \text{ for some } l \ge 0 \end{cases}$$ so $\|A - A_k\| = e^{-k} \to 0$.

For $j \in \mathbb{N}$ we have $$A^je_n = \alpha_n\alpha_{n+1}\cdots\alpha_{n+j-1}e_{n+j}$$ Notice that $$\alpha_{1}\alpha_2\cdots\alpha_{2^t-1} = \prod_{r=1}^{t-1} \exp(-r2^{t-r-1})$$

$$r(A)= \limsup_{j\to\infty}\|A^j\|^{\frac1j} \ge \limsup_{t\to\infty} \|A^{2^t-1}\|^{\frac1{2^t-1}} \ge \limsup_{t\to\infty} \|A^{2^t-1}e_1\|_2^{\frac1{2^{t-1}}} = \limsup_{t\to\infty} |\alpha_{1}\alpha_2\cdots\alpha_{2^t-1}|^{\frac1{2^{t-1}}} \\ \ge\limsup_{t\to\infty} \left(\prod_{r=1}^{t-1} \exp(-r2^{t-r-1})\right)^{\frac1{2^{t-1}}} = \limsup_{t\to\infty}\left(\prod_{r=1}^{t-1} \exp\left(-\frac{r}{2^{r}}\right)\right) = \limsup_{t\to\infty}\exp\left(-\sum_{r=1}^{t-1}\frac{r}{2^{r}}\right) = e^{-\sum_{r=1}^\infty \frac{r}{2^{r}}}$$

Therefore $A_k \to A$ but $r(A_k) \not\to r(A)$ since $r(A_k) = 0$ but $r(A) > 0$.

I am not sure if it is true in general, but I believe it is true if $\sigma(x)$ is discrete. We can prove the following:

For all $x\in A$ if $U$ is an open set containing $\sigma(x)$, there exists a $\delta_U>0$ such that $\sigma(x+y)\subset U$ for all $y\in A$ with $\|y\|<\delta$.

Proof: This is Theorem 10.20 in Rudin's functional analysis. I will reproduce the proof for convenience. The function $\mathbb C\backslash \sigma(x)\ni\lambda\mapsto\|(\lambda e-x)^{-1}\|$ is continuous on the resolvent set. Furthermore we know that as $|\lambda|\to \infty$ we must have $\|(\lambda e-x)^{-1}\|\to 0$. Thus we have some finite $M$ such that $\|(\lambda e-x)^{-1}\|<M$ for all $\lambda\notin U$. Hence if $y$ satisfies $\|y\|<1/M$ and $\lambda \notin U$ we have $\lambda e-(x+y)$ is invertible. This follows because $$\lambda e-(x+y)=(\lambda e-x)(e-(\lambda e-x)^{-1}y)$$ and $\|(\lambda e-x)^{-1}y\|<1$, so invertible.

We can also prove a strengthening of this:

If $U$ is an open set containing a component of $\sigma(x)$ (for any $x\in A$) and $\lim x_n=x$, then $\sigma(x_n)\cap U\neq \emptyset$ for all $n$ sufficiently large.

Proof: If $\sigma(x)$ is connected this follows from the above, so let us assume that it isn't. Thus there exists an open $V$ such that $\sigma(x)\subset U\cup V$. Assume by way of contradiction that the statement is false. Then for all $n\in \mathbb N$ there exists a $N\geq n$ such that $\sigma(x_n)\cap U=\emptyset$. In conjunction with the above theorem we thus have arbitrarily large $n$ such that $\sigma(x_n)\subset V$. As these $n$ are arbitrarily large we must be able to find a $x_n$, such that $\sigma(x_n)\subset V$ and $\|x-x_n\|<\delta_U$. The above theorem then implies that $\sigma(x)\subset V$, a contradiction.

Using this last theorem it is a fairly simple corollary that if $x$ has a discrete spectrum and $x_n\to x$ we must have $\lim \rho(x_n)=\rho(x)$. In particular this holds true for compact operators on Banach spaces.